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The quantum Fourier transform measurement supplies a rational approximationwhere $r$ is the desired multiplicative order. HereIts reduced continued fraction convergent has denominator two, so continued-fraction recovery in quantum order finding proposes $r=2$. Direct order verification from divisors confirms it:HenceAs a check, the classical final step of the Shor algorithm gives

As printed, the claim is false: $70$ is 7-cyclic-divisible because its rotations are $70$ and $07=7$, but its digits are not all equal to $7$ and its length is not a multiple of $6$.

The standard rotation argument proves the likely intended statement. If an $L$-digit integer and all its rotations are divisible by $7$, then for each leading digit $a$,is divisible by $7$. The multiplicative order of $10$ modulo $7$ is $6$. If $6\nmid L$, then $7\nmid 10^L-1$, forcing every digit $a$ to be divisible by $7$, hence to be either $0$ or $7$. Thus the valid conclusion is:The counterexample shows why “all its digits are equal to 7” cannot replace the first alternative.

The multiplicative order $r$ makes the states $|a^k\mathrm{mod}N⟩$, $0\le k<r$, distinct and cyclic under $U_a$. ThereforeThus each $|{\psi }_s⟩$ is an eigenvector withThese are the Fourier eigenvectors of the cyclic modular-multiplication orbit.

If a prime number $p$ divides the Fermat number $F_n$, it is odd and $2^{2^n}\equiv -1(\mathrm{mod}p)$. The multiplicative order of $2$ divides $2^{n+1}$ but not $2^n$, so it is exactly $2^{n+1}$. Fermat's little theorem then gives $2^{n+1}\mid p-1$. For $n\ge 1$, every prime divisor is $1$ modulo $4$. A prime need not divide any Fermat number merely because it is $1$ modulo $4$: order $12$ excludes the prime $13$ from the whole family.